Error in computing a limit I was given the following limit to compute,
$$\lim_{k\to\infty}\left[k-\sqrt{k^{2}+1}\right]$$
My approach:
$$= \lim_{k\to\infty}\left[k-\sqrt{k^{2}(1+k^{-2})}\right]$$
$$= \lim_{k\to\infty}[k-k]=0$$
So the following evaluates to $0$. But my book gives that answer is $\frac{-1}{2}$. Where did my process go wrong?
 A: $k-\sqrt{k^{2}(1+k^{-2})}$ is not equal to $k-k$. Hence you can't justify without further argument that
$$\lim_{k\to\infty}k-\sqrt{k^{2}(1+k^{-2})}=\lim_{k\to\infty}k-k=0.$$ What is true is that
$$k-\sqrt{k^{2}+1}= -\frac{1}{k+\sqrt{k^{2}+1}}$$ and that indeed the right hand side goes to zero as $k \to \infty$.
So your conclusion is right, but not the argument you gave.
A: $\require{\cancel}$
Your book is incorrect. Since $(a-b)(a+b) = a^2-b^2$ we get for $k>0$
\begin{align}
k - \sqrt{k^2 +1}& = k - \sqrt{k^2 +1} \left(\frac{k + \sqrt{k^2 +1}}{k + \sqrt{k^2 +1}} \right) \\
&= \frac{\cancel{k^2} - (\cancel{k^2} +1)}{k + \sqrt{k^2 +1}} \\
&= - \frac{1}{k + k\sqrt{ 1+\frac{1}{k^2}}} \\
& = -\frac{1}{k}\cdot\frac{1}{1+\sqrt{ 1+\frac{1}{k^2}}}
\end{align}
And since both $\lim_{k \to \infty}-\frac{1}{k} =0$ and $\lim_{k \to \infty}\frac{1}{1+\sqrt{ 1+\frac{1}{k^2}}} = \frac{1}{2}$ exist, your original limit is the product of these limits. Hence
$$
\lim_{k \to \infty}k - \sqrt{k^2 +1} =0\cdot \frac12 =0
$$
A: The book is wrong and you are right.
My approach is that $\displaystyle \left(k + \frac{1}{2k}\right)^2 > k^2 + 1.$
Therefore, $\displaystyle k - \sqrt{k^2 + 1} > k - \left(k + \frac{1}{2k}\right).$
So, as $k \to \infty,$ since $k - \left(k + \frac{1}{2k}\right) \to 0,$
so does $k - \sqrt{k^2 + 1}.$
A: Another standard approach is to use the Binomial expansion. For large positive $k,$
$$k-\sqrt{k^{2}+1}= k-\sqrt{k^{2}\left(1+k^{-2}\right)} = k-k\left(1+k^{-2}\right)^{1/2}$$
$$ = k - k\left[1+ \left( \frac{1}{2}\right) \frac{1}{k^2} + \frac{\left( \frac{1}{2}\right)\left( - \frac{1}{2}\right)}{2!} \left(\frac{1}{k^2}\right)^2 + \frac{\left( \frac{1}{2}\right)\left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)}{3!} \left(\frac{1}{k^2}\right)^3 + \ldots \right]\quad (1)$$
With a little bit of manipulation one sees, for example, that
$$ -\frac{1}{k} \left(\frac{1}{1-k^2} \right) = -\frac{1}{k} \sum_{n=0}^{\infty} \left(\frac{1}{k^2}\right)^n < (1) < 0$$
and by letting $k\to\infty$, The Sandwich Theorem gives us that expression $(1)\to 0.$
A: Your conclusion is correct. Another approach a bit over complicated for this problem is using a series expansion around $+\infty$, that is
$$x-(x^{2}+1)^{1/2}=-\frac{1}{2x}+\frac{1}{8x^{3}}+\mathcal{O}\left(\frac{1}{x^{5}}\right)\longrightarrow 0,\quad x\to +\infty.$$
